Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
    public static void main(String [] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        double P [] = new double[N+1];
        double DP[][] = new double[N+1][N+1];
        for(int i=1;i<=N;i++){
            P[i]= sc.nextDouble();
            Arrays.fill(DP[i],0);
        }
        DP[0][0]=1;
        for(int i=1;i<=N;i++){
            for(int j=0;j<=i;j++){
                if(j>0)DP[i][j] = DP[i-1][j-1] * P[i]+DP[i-1][j]*(1-P[i]);
                else DP[i][j]=DP[i-1][j]*(1-P[i]);
            }
        }
        double opt = 0;
        for(int i=(N+1)/2;i<=N;i++) opt += DP[N][i];
        System.out.println(opt);
    }
}